Find the coefficient of $x^{70}$ in the expansion of
\[(x - 1)(x^2 - 2)(x^3 - 3) \dotsm (x^{11} - 11)(x^{12} - 12).\]
The degree of the polynomial is $1 + 2 + 3 + \dots + 12 = \frac{12 \cdot 13}{2} = 78.$

When we expand $(x - 1)(x^2 - 2)(x^3 - 3) \dotsm (x^{11} - 11)(x^{12} - 12),$ we choose a term from each factor.  For example, from the first factor $x - 1,$ we can choose either $x$ or $-1.$  From the second factor $x^2 - 2,$ we can choose either $x^2$ or $-2,$ and so on.  So to find the coefficient of $x^{70},$ we want to cover all possible choices where the powers of $x$ multiply to $x^{70}.$

Since the degree of the polynomial is $x^{78},$ the product of the "missing" powers of $x$ must be $x^8.$  We divide into cases.

Case 1: One factor has a missing power of $x.$

If one factor has a missing power of $x,$ it must be $x^8 - 8,$ where we choose $-8$ instead of $x^8.$  Thus, this case contributes $-8x^{70}.$

Case 2: Two factors have a missing power of $x.$

If there are two missing powers of $x,$ then they must be $x^a$ and $x^b,$ where $a + b = 8.$   The possible pairs $(a,b)$ are $(1,7),$ $(2,6),$ and $(3,5)$ (note that order does not matter), so this case contributes $[(-1)(-7) + (-2)(-6) + (-3)(-5)] x^{70} = 34x^{70}.$

Case 3: Three factors have a missing power of $x.$

If there are three missing powers of $x,$ then they must be $x^a,$ $x^b,$ and $x^c,$ where $a + b + c = 8.$   The only possible triples $(a,b,c)$ are $(1,2,5)$ and $(1,3,4),$ so this case contributes $[(-1)(-2)(-5) + (-1)(-3)(-4)] x^{70} = -22x^{70}.$

Case 4: Four factors or more have a missing power of $x.$

If there are four or more missing powers of $x,$ then they must be $x^a,$ $x^b,$ $x^c,$ and $x^d$ where $a + b + c + d = 8.$  Since $a,$ $b,$ $c,$ $d$ are distinct, we must have $a + b + c + d \ge 10.$  Therefore, there are no ways to get a power of $x^{70}$ in this case.

Thus, the coefficient of $x^{70}$ is $(-8) + 34 + (-22) = \boxed{4}.$